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36. In a Young's double slit experiment the distance between the slits is 1.2 mmand the screen is at 0.75 m from the slits. If the distance of the 5th bright fringefrom the central fringe on the screen is 1.8 mm. Calculate the wavelength ofthe light used. What will be the distance of the 5th dark fringe from the centre ofthe screen? |
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Answer» Answer: ANSWER d=0.28mm=0.28\TIMES { 10 }^{ -3 }md=0.28mm=0.28×10 −3 m D=1.4mD=1.4m { x }_{ N }=1.35cm=1.35\times { 10 }^{ -2 }mx n
=1.35cm=1.35×10 −2 m n=5n=5 For n^{th}n th DARK fringe, { x }_{ n }=\cfrac { (2n-1) }{ 2 } \cfrac { \lambda D }{ d }x n
= 2 (2n−1)
d λD
1.35\times { 10 }^{ -2 }=\cfrac { \left[ \left( 2\times 5 \RIGHT) -1 \right] }{ 2 } \times \cfrac { \lambda \times 1.4 }{ 0.28\times { 10 }^{ -3 } }1.35×10 −2 = 2 [(2×5)−1]
× 0.28×10 −3
λ×1.4
\lambda =\cfrac { 1.35\times { 10 }^{ -2 }\times 0.28\times { 10 }^{ -3 }\times 2 }{ 1.4\times \left[ \left( 2\times 5 \right) -1 \right] } =0.06\times { 10 }^{ -5 }=6000Aλ= 1.4×[(2×5)−1] 1.35×10 −2 ×0.28×10 −3 ×2
=0.06×10 −5 =6000A When D=1.4-0.4m=1mD=1.4−0.4m=1m \beta =\cfrac { \lambda D }{ d } =\cfrac { 6000\times { 10 }^{ -10 }\times 1 }{ 0.28\times { 10 }^{ -3 } } =21428.5714\times { 10 }^{ -7 }=0.21cm=2.1429mmβ= d λD
= 0.28×10 −3
6000×10 −10 ×1
=21428.5714×10 −7 =0.21cm=2.1429mm |
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